# Simple Velocity question

1. Jun 17, 2008

### ritwik06

1. The problem statement, all variables and given/known data

The instantaneous velocity of a parrticle is given by the expression:
$$v=u sin \omega t$$
The time of complete cycle is $$T= \frac{2 \pi}{\omega}$$

Find out the average speed!

3. The attempt at a solution
After integration;
i get the following answer:
$$\frac{2u} { \pi}$$

tell me if I am wrong I will show my working if I am wrong.
Thanks a lot!

2. Jun 17, 2008

### armis

Hi ritwik

The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?

3. Jun 17, 2008

### ritwik06

Lol!! Thats so clever armis. But whatever your method might be, the answer will be same. So what did u get?
Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.

4. Jun 17, 2008

### alphysicist

Hi ritwik06,

Your result looks right to me.

5. Jun 17, 2008

### jablonsky27

ritwik, your result is wrong.
what limits did you take for integrating?

6. Jun 17, 2008

### armis

Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then $$\frac{2u}{\pi}$$ is the answer

Sorry, I didn't mean that

7. Jun 17, 2008

### alphysicist

Hi jablonsky27,

I got the same answer as ritwik. Are you integrating the correct function (they ask for the average speed)?

8. Jun 17, 2008

### jablonsky27

hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,

Vaverage = (1/T).$$\int$$(u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/$$\Pi$$T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex.

9. Jun 17, 2008

### ritwik06

Well if you go back and read the question, its written in bold face:
average speed

10. Jun 17, 2008

### ritwik06

Hey, I wasnt laughing at you dear. I was just wonering how u could take that out without integration. I mean the average speed. Anyways, thanks a lot for the effort.

11. Jun 17, 2008

### alphysicist

jablonsky,

I see ritwik already remarked about how we are looking for the average speed (not average velocity). So the integral to calculate would be:

$$\mbox{average speed }= \frac{1}{T}\ \int\limits_0^Tdt\ |u \sin(\omega t)|$$

Or you could restrict the integral to just one-fourth of a period.

12. Jun 17, 2008

### jablonsky27

oh ya. sorry. should learn to read questions more carefully in the future.